!Lecture #7& 8.
Definition 1. The conditional probability of an event A, given that an event B has occurred, is equal to
P(A |
B) = P(A∩B)/ P(B)
P(A∩B)=
P(A) P(B)
P(A |
B)= P(A) or P(B |
A)= P(B)
|
P(A∩B)
= P(A) P(B
| A) = P(B) P(A
| B) |
If
A1, A2,A3, …
form a sequence of pairwise mutually exclusive and exhaustive events in S (Ai
∩ Aj = Ø , i
≠j ) then
|
P(B)
= ∑ P(B|Ai)
P(Ai)
i |
|
P(Aj |B)
= [P(B|Aj)
P(Aj)]/∑ P(B|Ai)
P(Ai)
i |
j
= 1,2,…,k with P(Aj)
> 0 and P(B) > 0
SHORTCUT
IF A PROBLEM INVOLVES
TWO
EVENTS A and B
|
|
A |
A′ |
|
|
B |
P(A
∩ B) |
P(A′∩B) |
P(B) |
|
B′ |
P(A
∩ B′) |
P(A′∩
B′ ) |
P(B′) |
|
|
P(A) |
P(A′) |
1 |
|
|
|
|
|
Note:
P(A)
+ P(A′) = 1
P(B) + P(B′) = 1
P(A
∩ B) + P(A′∩B) = P(B)
P(A ∩ B′) + P(A′∩ B′ ) = P(B′)
P(A
∩ B) + P(A ∩ B′) = P(A)
P(A′∩B) + P(A′∩ B′ ) =P(A′)
The
event-composition approach to calculate the probability of an event of interest,
say event A, expresses A as a composition (unions and/or intersections) of two
or more other events. Then the two laws of probability can be applied to the
composition to find P(A). Mutually exclusive events simplify the addition law
and independence simplifies the multiplication law of probability.
The best way to learn how to solve probability
problems is to learn by doing!
PROBABILITY OF A COMPOSITION OF TWO OR MORE EVENTS. EXAMPLES.
Example1. In how many ways can 7 students line up
outside Statistics Professor’s door to complain about grades?
Example 2. A box contains 4 red balls and 6 other
balls. Two balls are drawn
without replacement. What is the probability that
they are red?
Example 3. A box contains 8 red balls and 14 black
balls. Three balls are drawn
without replacement. (a) What is the probability
that all 3 are red?
(b) What is the probability that 1 is
red and 2 are black?
Example 4. In a certain community 30% of the
people smoke, 55% of them drink alcohol, and 20% of them smoke as well as drink.
Calculate the probability that a randomly selected person
(i)
smokes but does not drink
(ii)
neither smokes nor drink
(iii)
either smokes or does not drink or both
Example 5 . A
bank has five junior executive in the head branch. Each year, one of these five
is selected at random to be transferred as
manager to one of the local branches. The selected individual is replaced by a
new junior executive. Find the probability that a given junior executive:
(i)
Stays exactly 2 years in the head branch
(ii)
Stays more than 2 years in the head branch
(iii)
Stays 3 years or less in the head branch.
Example 6
. In
a large population of fruit flies, 30% of the flies have a wing mutation, 40%
have an eye mutation, and 15% have both eye and wing mutations What is the
probability that a fly has at least one of the mutations?
Example 7c .
A basket contains 2 red balls and 2 white balls. A ball is drawn and set
aside, and its color is noted. Then a second ball is drawn. What is the
probability that both are red?
Example 8ci. A card is drawn from a deck of cards.
Then the card is replaced, the deck is reshuffled, and a second card is drawn.
What is the probability of getting
an ace on the first and a king on the second?
Example 9c.The experiment involves the selection
of 2 applicants out of 5. Find the
probability of drawing exactly one of the two best
applicants, event A.
Example 10c. A monkey is to be taught to recognize
colors by tossing one red, one
black, and one white ball into boxes of the same
respective colors, one ball to a box.
If the monkey has not learned the colors, and
merely tosses one ball into each box at
random, find the probability of
(a) No color matches. (b)Exactly one color match.
Example
11b. In answering a question on a multiple choice test a student either knows
the answer or he guesses. Let p be the probability that he knows the answer.
Assume that a student who guesses at the answer will be correct with probability
1/m , where m is the number of multiple choice alternatives. What is the
conditional probability that a student knew the answer to a question given that
he answered it correctly?
Example 12b. An oil company estimates that from
geological data there is a probability of 0.3 of finding oil in a certain area.
It knows from previous experience that if oil is to be found, there is a
probability of 0.4 that a positive strike of some kind will be made on the first
series of drillings. If the first series of drillings turn out to be negative,
what is the probability that oil will eventually be found? Suppose that a second
drillings turn out to be negative. What then will be the new probability of
eventually striking oil, if (a) assuming that it knows from previous experience
that every series of strikes is
independent; (b) assuming
that it knows from previous experience that if oil is to be found, there is a
probability of 0.6 that a positive strike of some kind will be made on the
second series of drillings?